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Version 2 |
| Let $G$ be a locally compact Hausdorff topological group and $\mu$ a left Haar measure. Although left and right Haar measures in $G$ always exist, they generally do not coincide, i.e. a left Haar measure is usually not invariant under right translations. Nevertheless, the right translations of a left Haar measure can be easily described as explained in the following theorem. |
Let $G$ be a locally compact Hausdorff topological group and $\mu$ a left Haar measure. Although left and right Haar measures in $G$ always exist, they generally do not coincide, i.e. a left Haar measure is usually not invariant under right translations. Nevertheless, the right translations of a left Haar measure can be easily described as explained in the following theorem. |
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| {\bf Theorem -} Let $G$ be a locally compact Hausdorff topological group and $\mu$ a left Haar measure in $G$. Then, there exists a continuous homomorphism $\Delta:G \longrightarrow \mathbb{R}^+$ such that, for every $t \in G$ and every measurable subset $A$ |
{\bf Theorem -} Let $G$ be a locally compact Hausdorff topological group and $\mu$ a left Haar measure in $G$. Then, there exists a continuous homomorphism $\Delta:G \longrightarrow \mathbb{R}^+$ such that, for every $t \in G$ and every measurable subset $A$ |
| \begin{displaymath} |
\begin{displaymath} |
| \mu(At) = \Delta(t^{-1})\mu(A) |
\mu(At) = \Delta(t^{-1})\mu(A) |
| \end{displaymath} |
\end{displaymath} |
| Moreover, if $f:G \longrightarrow \mathbb{C}$ is an integrable function then |
Moreover, if $f:G \longrightarrow \mathbb{C}$ is an integrable function then |
| \begin{displaymath} |
\begin{displaymath} |
| \Delta(t)\int_Gf(st) \mu(s) = \int_G f(s) \mu(s) |
\Delta(t)\int_Gf(st) \mu(s) = \int_G f(s) \mu(s) |
| \end{displaymath} |
\end{displaymath} |
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| The function $\Delta$ is called the {\bf modular function} of $G$ (notice that, by uniqueness up to scalar multiple of left Haar measures, $\Delta$ only depends on $G$). Other names for $\Delta$ that can be found are: \emph{Haar modulus}, or \emph{modular character} or \emph{modular homomorphism}. |
The function $\Delta$ is called the {\bf modular function} of $G$ (notice that, by uniqueness up to scalar multiple of left Haar measures, $\Delta$ only depends on $G$). Other names for $\Delta$ that can be found are: \emph{Haar modulus}, or \emph{modular character} or \emph{modular homomorphism}. |
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We now prove the above theorem, except the the continuity of $\Delta$ (which is slightly harder to obtain).
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We now prove the above theorem, except the the continuity of $\Delta$ (which is slightly harder to obtain).
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| {\bf \emph{Proof (execpt continuity of $\Delta$):}} |
{\bf \emph{Proof (execpt continuity of $\Delta$):}} |
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| Let $t \in G$. The function $\nu$, defined on measurable subsets $A$ by |
Let $t \in G$. The function $\nu$, defined on measurable subsets $A$ by |
| \begin{displaymath} |
\begin{displaymath} |
| \nu (A):= \mu(At) |
\nu (A):= \mu(At) |
| \end{displaymath} |
\end{displaymath} |
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| is easily seen to be a measure in $G$. Moreover, $\nu$ is left invariant (since $\mu$ is left invariant). By the uniqueness of left Haar measures, $\mu$ must be a multiple of $\nu$, i.e. $\mu=\Delta(t)\nu$ for some positive scalar $\Delta(t) \in \mathbb{R}^+$. Thus, we have proven that for every measurable subset $A$ |
is easily seen to be a measure in $G$. Moreover, $\nu$ is left invariant (since $\mu$ is left invariant). By the uniqueness of left Haar measures, $\mu$ must be a multiple of $\nu$, i.e. $\mu=\Delta(t)\nu$ for some positive scalar $\Delta(t) \in \mathbb{R}^+$. Thus, we have proven that for every measurable subset $A$ |
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| \begin{displaymath} |
\begin{displaymath} |
| \mu(At)= \Delta(t)^{-1}\mu(A) |
\mu(At)= \Delta(t)^{-1}\mu(A) |
| \end{displaymath} |
\end{displaymath} |
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| Now for $s, t \in G$ we have that $\mu(Ast)=\Delta(st)^{-1}\mu(A)$, but also |
Now for $s, t \in G$ we have that $\mu(Ast)=\Delta(st)^{-1}\mu(A)$, but also |
| \begin{itemize} |
\begin{itemize} |
| \item $\mu(Ast)= \Delta(t)^{-1}\mu(As)$, and |
\item $\mu(Ast)= \Delta(t)^{-1}\mu(As)$, and |
| \item $\mu(As) = \Delta(s)^{-1}\mu(A)$ |
\item $\mu(As) = \Delta(s)^{-1}\mu(A)$ |
| \end{itemize} |
\end{itemize} |
| So, we can see that, for every measurable subset $A$, |
So, we can see that, for every measurable subset $A$, |
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| \begin{displaymath} |
\begin{displaymath} |
| \Delta(st)^{-1}\mu(A) = \Delta(t)^{-1}\Delta(s)^{-1}\mu(A) |
\Delta(st)^{-1}\mu(A) = \Delta(t)^{-1}\Delta(s)^{-1}\mu(A) |
| \end{displaymath} |
\end{displaymath} |
| Hence, $\Delta(st) = \Delta(t)\Delta(s)$. Thus, $\Delta$ is an homomorphism. |
Hence, $\Delta(st) = \Delta(t)\Delta(s)$. Thus, $\Delta$ is an homomorphism. |
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| The statement about integrals of functions follows easily by approximation by simple functions. For simple functions it is easy to see it is true using the now established condition $\mu(At) = \Delta(t^{-1})\mu(A)$. $\square$ |
The statement about integrals of functions follows easily by approximation by simple functions. For simple functions it is easy to see it is true using the now established condition $\mu(At) = \Delta(t^{-1})\mu(A)$. $\square$ |
